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Quadratic Equation and Inequalities question

2015 · Shift 0 · Q26
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  5. /2015 · Shift 0 · Q26

Quadratic Equation and Inequalities question

2015 · Shift 0 · Q26

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α\alphaα and β\betaβ be the roots of equation x2−6x−2=0{x^2} - 6x - 2 = 0x2−6x−2=0. If an=αn−βn,{a_n} = {\alpha ^n} - {\beta ^n},an​=αn−βn, for n≥1,n \ge 1,n≥1, then the value of a10−2a82a9{{{a_{10}} - 2{a_8}} \over {2{a_9}}}2a9​a10​−2a8​​ is equal to :
  1. A
    333
  2. B
    −3- 3−3
  3. C
    666
  4. D
    −6- 6−6
View written solutionFree

Correct answer: A

  1. Given quadratic and its roots

The roots α,β\alpha, \betaα,β satisfy x2−6x−2=0.x^2-6x-2=0.x2−6x−2=0. So, α+β=6,αβ=−2.\alpha+\beta=6, \qquad \alpha\beta=-2.α+β=6,αβ=−2.

Also, since α\alphaα is a root, α2=6α+2,\alpha^2=6\alpha+2,α2=6α+2, and similarly, β2=6β+2.\beta^2=6\beta+2.β2=6β+2.

  1. Given sequence

an=αn−βn.a_n=\alpha^n-\beta^n.an​=αn−βn. We need to find a10−2a82a9.\frac{a_{10}-2a_8}{2a_9}.2a9​a10​−2a8​​.

  1. Find a recurrence for ana_nan​

From α2=6α+2,\alpha^2=6\alpha+2,α2=6α+2, multiply by αn−2\alpha^{n-2}αn−2: αn=6αn−1+2αn−2.\alpha^n=6\alpha^{n-1}+2\alpha^{n-2}.αn=6αn−1+2αn−2. Similarly, βn=6βn−1+2βn−2.\beta^n=6\beta^{n-1}+2\beta^{n-2}.βn=6βn−1+2βn−2. Subtracting, an=6an−1+2an−2.a_n=6a_{n-1}+2a_{n-2}.an​=6an−1​+2an−2​.

Thus, a10=6a9+2a8.a_{10}=6a_9+2a_8.a10​=6a9​+2a8​.

So, a10−2a8=(6a9+2a8)−2a8=6a9.a_{10}-2a_8=(6a_9+2a_8)-2a_8=6a_9.a10​−2a8​=(6a9​+2a8​)−2a8​=6a9​.

Hence, a10−2a82a9=6a92a9=3,\frac{a_{10}-2a_8}{2a_9}=\frac{6a_9}{2a_9}=3,2a9​a10​−2a8​​=2a9​6a9​​=3, provided a9≠0a_9\neq 0a9​=0.

Since α≠β\alpha\neq \betaα=β, clearly a9=α9−β9≠0a_9=\alpha^9-\beta^9\neq 0a9​=α9−β9=0.

  1. Evaluate options

The value is 3.3.3. So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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